Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $k = \dfrac{-5a^2 + 15a + 20}{-5a^3 + 25a^2 + 30a} \times \dfrac{a^2 + 2a}{a - 4} $
Explanation: First factor out any common factors. $k = \dfrac{-5(a^2 - 3a - 4)}{-5a(a^2 - 5a - 6)} \times \dfrac{a(a + 2)}{a - 4} $ Then factor the quadratic expressions. $k = \dfrac {-5(a + 1)(a - 4)} {-5a(a + 1)(a - 6)} \times \dfrac {a(a + 2)} {a - 4} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac { -5(a + 1)(a - 4) \times a(a + 2)} { -5a(a + 1)(a - 6) \times (a - 4)} $ $k = \dfrac {-5a(a + 1)(a - 4)(a + 2)} {-5a(a + 1)(a - 6)(a - 4)} $ Notice that $(a + 1)$ and $(a - 4)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {-5a\cancel{(a + 1)}(a - 4)(a + 2)} {-5a\cancel{(a + 1)}(a - 6)(a - 4)} $ We are dividing by $a + 1$ , so $a + 1 \neq 0$ Therefore, $a \neq -1$ $k = \dfrac {-5a\cancel{(a + 1)}\cancel{(a - 4)}(a + 2)} {-5a\cancel{(a + 1)}(a - 6)\cancel{(a - 4)}} $ We are dividing by $a - 4$ , so $a - 4 \neq 0$ Therefore, $a \neq 4$ $k = \dfrac {-5a(a + 2)} {-5a(a - 6)} $ $ k = \dfrac{a + 2}{a - 6}; a \neq -1; a \neq 4 $